Question

How do you solve `3x^2-16x+5<=0` by algebraically?

Answer 1

The solution is `x in [1/3, 5]`

Explanation:

First factorise the equation

`3x^2-16x+5=(3x-1)(x-5)<=0`

`(3x-1)(x-5)=0`

when

`{(3x-1=0),(x-5=0):}`

`{(x=1/3),(x=5):}`

Let `f(x)=(3x-1)(x-5)`

Build a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaaaaa)` `1/3` `color(white)(aaaaaaaa)` `5` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `3x-1` `color(white)(aaaa)` `-` `color(white)(aaaa)` `0` `color(white)(aaaa)` `+` `color(white)(aaaaa)` `+`

`color(white)(aaaa)` `x-5` `color(white)(aaaaa)` `-` `color(white)(aaaa)` ###color(white)(aaaaa)#`-` `color(white)(aa)` `0` `color(white)(aa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaa)` `+` `color(white)(aaaa)` `0` `color(white)(aaaa)` `-` `color(white)(aa)` `0` `color(white)(aa)` `+`

Therefore,

`f(x)<=0` when `x in [1/3, 5]`